Topic 9 Oxidation and Reduction. IB Core Objective 9.1.1 Define oxidation and reduction in terms of electron loss and gain. Define: Give the precise meaning

Topic 9

Oxidation and Reduction

IB Core Objective

• 9.1.1 Define oxidation and reduction in terms of electron loss and gain.

• Define: Give the precise meaning of a word, phrase or physical quantity.

9.1.1 Define oxidation and reduction in terms of electron loss and gain.

Oxidation: The loss of electronsFe2+(aq) → Fe3+(aq) + e-

Reduction: The gain of electrons2H+(aq) + 2e- → H2(g)


Helpful Mnemonic

This is Leo the Lion

LEO goes GER

Loss of Electrons is Oxidation

Gain of Electrons is Reduction


Or another if you prefer…

OIL RIGOxidation Is Loss of

electrons.Reduction Is Gain of

electrons.

IB Core Objective

• 9.1.2 Deduce the oxidation number of an element in a compound.

• Deduce: Reach a conclusion from the information given.

9.1.2 Deduce the oxidation number of an element in a compound.

In order to keep track of what loses electrons and what gains them, we assign oxidation numbers.


A species is oxidized when it loses electrons.– Here, zinc loses two electrons to go from neutral

zinc metal to the Zn2+ ion.


A species is reduced when it gains electrons.– Here, each of the H+ gains an electron and they

combine to form H2.


• It may be easier to find what is being reduced and oxidized by splitting the equation into “half equations”.

• For example, with Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g)

It can be split up as:Zn(s) → Zn2+(aq) + 2e-

and 2H+(aq) + 2e- → H2(g)


• It is not always easy to split equations into half equations.

• Consider the following reaction:

Can you tell which is being oxidized? If not, then we need to use oxidation numbers.

N2(g) + 3H2(g) 2NH3(g)


Oxidation NumberThe charge that an atom would have if all

covalent bonds were broken so that the more electronegative element kept all the electrons.


Oxidation Number Rules– Elements in elemental state = 0– F = -1 (always)– O = -2 (except in H2O2 where its +1)– H = +1 (except in hydrides H-)– Halides = -1 except when bonded to oxygen or other halides

higher in the group (more reactive one will be -1)

The sum of the oxidation numbers in a neutral compound is 0.The sum of the oxidation numbers in a polyatomic ion is the

charge on the ion.


Find the oxidation number for the following:

Nitrogen in N2 =

Carbon in CH4 =

Sulfur in H2SO4 =

Phosphorous in PCl4+ =

Iodine in IO4- =

Answers: 0, -4, +6, +5, +7

– Elements in elemental state = 0

– F = -1 (always)

– O = -2 (except in H2O2 where its +1)

– H = +1 (except in hydrides H-)

– Halides = -1 except when bonded to oxygen or other halides higher in the group (more reactive one will be -1)

IB Core Objective

• 9.1.4 Deduce whether an element undergoes oxidation or reduction in reactions using oxidation numbers.

• Deduce: Reach a conclusion from the information given. (Obj 3)

9.1.4 Deduce whether an element undergoes oxidation or reduction in reactions using oxidation numbers.

Let’s go back to the equation:

What is the oxidation number for nitrogen on both sides?

Has it been oxidized or reduced?

Answer: Oxidation number goes from 0 to -3. It has gained electrons, therefore it has been reduced.

N2(g) + 3H2(g) 2NH3(g)


Consider the reaction between MnO4− and C2O4

2− :

MnO4−(aq) + C2O4

2−(aq) Mn2+(aq) + CO2(aq)


MnO4− + C2O4

2- Mn2+ + CO2

First, assign oxidation numbers.

+7 +3 +4+2

Since the manganese goes from +7 to +2, it is reduced.

Since the carbon goes from +3 to +4, it is oxidized.

Next, find out if carbon and manganese are being oxidized or reduced.

IB Core Objective

• 9.1.3 State the names of compounds using oxidation numbers.

• State: Give a specific name, value or other brief answer without explanation or calculation. (Obj 1)

9.1.3 State the names of compounds using oxidation numbers.

For elements that have a variable oxidation number, the oxidation state is signified by Roman numerals.

Example: Fe+3 would be written as Iron(III)How would you write the following?FeCl2 FeCl3 MnO4

- Cr2O3

Answers: iron(II) chloride, iron(III) chloride, permanganate (VII), chromium(III) oxide

Challenge: How would you write the formula for ammonium dichromate?

Answer: (NH4)2Cr2O7

Ammonium dichromate volcano

• (NH4)2Cr2O7 --> Cr2O3 + 4 H2O + N2

• Is chromium oxidized or reduced in this reaction?

• Is nitrogen oxidized or reduced in this reaction?Answer: Chromium is reduced from +6 to +3

Nitrogen is oxidized from +3 to 0

IB Core Objective

• 9.2.1 Deduce simple oxidation and reduction half-equations given the species involved in a redox reaction.


9.2.1 Deduce simple oxidation and reduction half-equations given the species involved in a redox reaction.

• Let’s look at an equation that we worked with before….

• What is wrong with this equation?• Answer: It is not balanced!• We have worked with half equations before

(zinc and hydrogen). Now we’ll dig deeper.

MnO4− + C2O4

2- Mn2+ + CO2


General rules for balancing half equations

– 1) Balance atoms being oxidized or reduced– 2) Add H20 to balance Oxygen atoms

– 3) Add H+(aq) to balance Hydrogen atoms

– 4) Add e- to balance charge


Oxidation Half-Reaction

C2O42− CO2

To balance the carbon, we add a coefficient of 2:

C2O42− 2 CO2


Oxidation Half-ReactionC2O4

2− 2 CO2

The oxygen is now balanced as well. To balance the charge, we must add 2 electrons to the right side.

C2O42− 2 CO2 + 2 e−


Reduction Half-Reaction

MnO4− Mn2+

The manganese is balanced; to balance the oxygen, we must add 4 waters to the right side.

MnO4− Mn2+ + 4 H2O



MnO4− Mn2+ + 4 H2O

To balance the hydrogen, we add 8 H+ to the left side.

8 H+ + MnO4− Mn2+ + 4 H2O



8 H+ + MnO4− Mn2+ + 4 H2O

To balance the charge, we add 5 e− to the left side.

5 e− + 8 H+ + MnO4− Mn2+ + 4 H2O

IB Core Objective

• 9.2.2 Deduce redox equations using half-equations.


9.2.2 Deduce redox equations using half-equations.

Combining the Half-ReactionsNow we evaluate the two half-reactions

together:C2O4

2− 2 CO2 + 2 e−

5 e− + 8 H+ + MnO4− Mn2+ + 4 H2O

To attain the same number of electrons on each side, we will multiply the first reaction by 5 and the second by 2.


Combining the Half-Reactions5 C2O4

2− 10 CO2 + 10 e−

10 e− + 16 H+ + 2 MnO4− 2 Mn2+ + 8 H2O

When we add these together, we get:

10 e− + 16 H+ + 2 MnO4− + 5 C2O4

2−

2 Mn2+ + 8 H2O + 10 CO2 +10 e−


Combining the Half-Reactions10 e− + 16 H+ + 2 MnO4

− + 5 C2O42−

2 Mn2+ + 8 H2O + 10 CO2 +10 e−

The only thing that appears on both sides are the electrons. Subtracting them, we are left with:

16 H+ + 2 MnO4− + 5 C2O4

2− 2 Mn2+ + 8 H2O + 10 CO2



PracticeGiven two half-equations:

Cr2O72-

(aq) → Cr3+(aq)

Fe2+ → Fe3+

Deduce the half-equations for each, then deduce the redox equation.

Answer

Cr2O72-

(aq) + 14H+(aq) + 6Fe2+

(aq) →

2Cr3+(aq) + 7H2O(l) + 6Fe3+

(aq)

IB Core Objective

• 9.2.3 Define the terms oxidizing agent and reducing agent.

• Define: Give the precise meaning of a word, phrase or physical quantity. (Obj 1)

9.2.3 Define the terms oxidizing agent and reducing agent.

Oxidizing agent: Substance that is reduced and causes the oxidation of another substance in a redox reaction.

Reducing agent: Substance that is oxidized and causes the reduction of another substance in a redox reaction.

I am oxidizing agent man.I am here to take your

electrons.

IB Core Objective

• 9.2.4 Identify the oxidizing and reducing agents in redox equations.

• Identify: Find an answer from a given number of possibilities. (Obj 2)

9.2.4 Identify the oxidizing and reducing agents in redox equations.

Identify the oxidizing and reducing agents in the following equations:

Sn2+(aq) + 2Fe3+

(aq) → Sn4+(aq) Fe2+

(aq)

Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)

9.2.4 Identify the oxidizing and reducing agents in redox equations.

Deduce the following half equations, deduce the redox equation, and identify the oxidizing agent and the reducing agent.

• MnO4-(aq) → Mn-2

(aq)

• SO2(aq) → SO42-

(aq)

IB Core Objective

• 9.3.1 Deduce a reactivity series based upon the chemical behaviour of a group of oxidizing and reducing agents.


9.3.1 Deduce a reactivity series based upon the chemical behaviour of a group of oxidizing and reducing agents.

• Recall in acids and bases that a strong acid had a weak conjugate base.

• Same in redox reactions. The conjugate of a powerful oxidizing agent is a weak reducing agent.

F2 + 2e- ↔ 2F-Strong oxidizing

agentWeak reducing

agent


• Mr. F can really attract the electrons (more electronegative).

• When Mr. F has the electrons, he doesn’t want to let them go.

• So although he is a good oxidizing agent, he is a poor reducing agent. (He doesn’t like to reduce the number of his electrons!)

9.3.1 Deduce a reactivity series based upon the chemical behaviour of a group of oxidizing and reducing

agents.

• Think back to Topic 3 on Periodicity.• What are the trends in electronegativity?


Compare

What exception do you see?

Hydrogen (Lithium is anotherexception)

IB Core Objective

• 9.3.2 Deduce the feasibility of a redox reaction from a given reactivity series.


9.3.2 Deduce the feasibility of a redox reaction from a given reactivity series.

Cl2(aq) + 2I-(aq) → I2(aq) + 2Cl-

(aq)

Feasible? A: Yes

I2(aq) + 2Cl-(aq) → Cl2(aq) + 2I-

(aq)

Feasible?

A: No

Chlorine attracts electrons more strongly than iodine, so chlorine is a better oxidizing agent.

9.3.2 Deduce the feasibility of a redox reaction from a given reactivity series.

Zn(s) + Cu2+(aq) → Cu(s) + Zn2+

(aq)

Feasible?A: Yes

Cu(s) + Zn2+(aq) → Zn(s) + Cu2+

(aq)

Feasible?A: No

These examples are all displacement reactions, because they involve a more reactive metal or non-

metal displacing the reactive one from its salt.

IB Core Objective

• 9.4.1 Explain how a redox reaction is used to produce electricity in a Voltaic cell.

• Explain: Give a detailed account of causes, reasons or mechanisms.

9.4.1 Explain how a redox reaction is used to produce electricity in a Voltaic cell.

A Voltaic cell is a device for converting chemical energy into electrical energy using a redox reaction.


• Anode(-): Oxidation, forms a negative charge• Cathode(+): Reduction, forms a positive charge

2+2+e

-e-e-

e-

e-e-

2+

2+

e-e-


• Lets harness some Energy!!

2+2+

2+

2+

2+

Zn(s) Cu(s)

2+2+

e-e-

2+

2+2+

e-e-

Problem, the highly negative charge on electrode causes (+) ions to be attracted

back

Solution

Balance (-) charge by replacing it

with some more negative ions


• Lets harness some Energy!!

2+

2+2+

2+

2+

Zn(s) Cu(s)

2+2+

e-e-

2+

2+2+

http://www.dynamicscience.com.au/tester/solutions/chemistry/redox/galvan5.swf

+-

+-

+

-

+

-+

-

e-e-

2+





IB Core Objective

• 9.4.2 State that oxidation occurs at the negative electrode (anode) and reduction occurs at the positive electrode (cathode).

• State: Give a specific name, value or other brief answer without explanation or calculation.

9.4.2 State that oxidation occurs at the negative electrode (anode) and reduction occurs at the positive

electrode (cathode).

• A typical cell looks like this.

• The oxidation occurs at the anode.

• The reduction occurs at the cathode.

• Which of the metals is being reduced?

• So which is the cathode?

9.4.2 State that oxidation occurs at the negative electrode (anode) and reduction occurs at the positive electrode (cathode).

• Lead and zinc are set up in a voltaic cell.• Which one would be oxidized? Which one is

being reduced?• A: Zinc is being oxidized. Lead is being

reduced. • Which one would be the cathode and which

would be the anode?• Zinc would be the anode, lead is the cathode.

IB Core Objective

• 9.5.1 Describe, using a diagram, the essential components of an electrolytic cell.

• Describe: Give a detailed account.

9.5.1 Describe, using a diagram, the essential components of an electrolytic cell.

Homework:• Draw a diagram of an electrolytic cell.• Provide a brief description what is happening at

each step, including the components, where oxidation and reduction is occurring, how current is conducted, and the products of a molten salt.

• If you do this effectively, you will have down objectives 9.5.1 – 9.5.4

9.5.1 Describe, using a diagram, the essential components of an electrolytic cell.

Need to have a liquid containing ions, which is called an electrolyte.

IB Core Objective

• 9.5.2 State that oxidation occurs at the positive electrode (anode) and reduction occurs at the negative electrode (cathode).

• State: Give a specific name, value or other brief answer without explanation or calculation.

9.5.2 State that oxidation occurs at the positive electrode (anode) and reduction occurs at the negative electrode (cathode).

• The anode attracts anions.• When the anions reach they anode, they lose

electrons.• So are they oxidized or reduced?• A: oxidized• When cations reach the cathode they gain

electrons and they are reduced.

IB Core Objective

• 9.5.3 Describe how current is conducted in an electrolytic cell.


9.5.3 Describe how current is conducted in an electrolytic cell.

• Electricity is supplied from an external source and used to create a non-spontaneous reaction.

• Electrolyte solution can conduct electricity because the ions move towards oppositely charged electrodes.

IB Core Objective

• 9.5.4 Deduce the products of the electrolysis of a molten salt.


9.5.4 Deduce the products of the electrolysis of a molten salt.

Sodium chloride• Negative chloride ions are attracted to the

positive ions. There they lose electrons and are oxidized to chlorine gas:

2Cl-(l) → Cl2(g) + 2e-

• Positive sodium ions are attracted to the negative cathode. They gain electrons and are reduced to sodium metal:

Na+(l) + e- → Na(l)

9.5.4 Deduce the products of the electrolysis of a molten salt.

QuestionFor every 2 mol of electrons that flow through

the circuit, how many mol of chlorine gas and sodium metal will be produced?

A: 1 mol of chlorine gas and 2 mol of sodium.

IB HL Objective

• 19.1.1 Describe the standard hydrogen electrode.

• Describe: Give a detailed account. (Obj 2)

19.1.1 Describe the standard hydrogen electrode.

Electrode: An electrical conductor through which electric current leaves or enters

Anode: Negative electrode where oxidation takes place.

Cathode: Positive electrode where reduction takes place.


• The potential of any two electrodes can be compared using this apparatus

• You will learn more about this voltaic cell later (SL topic).


• Electrode potentials from the voltaic cell are measured relative to the standard hydrogen electrode (SHE).


• The reference half-reaction is the reduction of H+(aq) to H2(g): 2H+(aq) + 2e- ↔ H2(g)

IB HL Objective

• 19.1.2 Define the term standard electrode potential (Eѳ).

• Define: Give the precise meaning of a word, phrase or physical quantity. (Obj 1)

Electromotive Force (emf)Water only

spontaneously flows one way in

a waterfall.

Likewise, electrons only spontaneously flow one way in a redox reaction—from higher to lower potential energy.

Electromotive Force (emf)• The potential difference between the anode

and cathode in a cell is called the electromotive force (emf).

• It is also called the cell potential, and is designated Ecell.

19.1.2 Define the term standard electrode potential (Eѳ).


• The standard hydrogen electrode (SHE) is defined as having a potential of zero.

• Standard electrode potentials also refer to the conditions. In the SHE, the platinum electrode is surrounded by H2 gas at 1 atm (1.01 x 105 Pa), electrode is immersed in strong acid at 1.00 mol dm-3, and is kept at 298 K.

• in a standard hydrogen electrode is equal to 0 V.• is the standard reduction potential.


The electrode potential at standard conditions can be found through this equation:

Ecell = Ered (cathode) − Ered (anode)

There is also another way….If the half-equation is being oxidized instead of reversed, just flip the values . Example: K ↔ K+ +e- Eѳ = +2.92 V.Then add the two values from the two half-reactions together!

IB HL Objective

• 19.1.3 Calculate cell potentials using standard electrode potentials.

• Calculate: Find a numerical answer showing the relevant stages in the working (unless instructed not to do so). (Obj 2)

19.1.3 Calculate cell potentials using standard electrode potentials.

Reduction potentials for

many electrodes have been

measured and tabulated.

Standard Electrode Potentials


• For the oxidation in this cell,

• For the reduction,

Ered = −0.76 V

Ered = +0.34 V


Ecell = Ered (cathode) − Ered (anode)

= +0.34 V − (−0.76 V)= +1.10 V

IB HL Objective

• 19.1.4 Predict whether a reaction will be spontaneous using standard electrode potential values.

• Predict: Give an expected result. (Obj 3)

19.1.4 Predict whether a reaction will be spontaneous using standard electrode potential values.

• Standard electrode potentials allow predictions to be made about which reactions could theoretically occur.

• If the cell potential is negative, a spontaneous reaction cannot occur.

• If the cell potential is positive, then the reaction could occur spontaneously.

19.1.4 Predict whether a reaction will be spontaneous using standard electrode potential values.

Can copper metal reduce hydrogen ions to hydrogen gas?

Cu ↔ Cu2+ + 2e- Eѳ = -0.34 V2H+ + 2e- ↔ H2 Eѳ = 0.00 V

Eѳcell = -0.34 V

Answer= No

IB HL Objective

• 19.2.1 Predict and explain the products of electrolysis of aqueous solutions.

• Predict: Give an expected result.• Explain: Give a detailed account of causes,

reasons or mechanisms.

19.2.1 Predict and explain the products of electrolysis of aqueous solutions.

QuestionHow would you know if a solution has electrolytes

in it?A: It conducts electricity.Electrolytes are easy to determine the products,

since there is one positive ion and one negative ion.

So sodium chloride may dissociate in solution. Anything else?


• Water is a poor conductor of electricity, but it still dissociates into ions:

H2O(l) ↔ H+(aq) + OH-

(aq)

In order to electrolyze water, need to add something more to conduct the current that easily produces ions, but won’t be oxidized/reduced. So a small amount of sulfuric acid is added.


• If water is electrolyzed, what would the products be? Half-equations for each ion?

• A: for the hydrogen ions:2H+(aq) + 2e- → H2(g)

for the hydroxide ions:4OH-(aq) → O2(g) + 2H2O(l) + 4e-


• You will also need to know aqueous sodium chloride for this objective.

• We will be going over this more with the SL students, since they will need to know this as well.


• What if we were to electrolyze copper(II) sulfate solution? What are equations for ions being formed in solution?

• A: CuSO4(aq) → Cu2+(aq) + SO42-(aq)

• H2O(l) ↔ H+(aq) + OH-(aq)

• Inert (non-reactive) platinum or graphite electrodes can be used.


• Based on the dissociation equations, which would be attracted to the positive electrode? Which would be attracted to the negative electrode? What are the half-reactions for these? Be sure to take into account Eѳ potentials.

(-) electrode: Cu2+(aq) + 2e- → Cu(s)(+) electrode: 4OH-(aq) → 2H2O(l) + O2(g) + 4e-

Because copper is below hydrogen in the reactivity series, it will gain electrons instead of the hydrogen. Because the sulfate would have a more positive value, it would keep its electrons over the hydroxide.


• What if a copper electrode is used? What would the half equations be at each electrode?

• (-) electrode: Cu2+(aq) + 2e- → Cu(s)• (+) electrode: Cu(s) → Cu2+(aq) + 2e-

IB HL Objective

• 19.2.2 Determine the relative amounts of the products formed during electrolysis.

• Determine: Find the only possible answer.

19.2.2 Determine the relative amounts of the products formed during electrolysis.

• In the electrolysis of water, what would the mol ratio be for products?

• For oxygen gas: 4OH-(aq) → O2(g) + 2H2O(l) + 4e-

We need four mol of electrons to produce one mol of oxygen.

• For hydrogen gas:2H+(aq) + 2e- → H2(g)

Four mol of electrons would produce two mol of hydrogen gas.

Therefore the ratio would be 2 mol H2: 1 mol O2

19.2.2 Determine the relative amounts of the products formed during electrolysis.

What factors will influence the amount of products produced and the rate?

1. The magnitude of current (increasing the flow of electrons). The stronger the current, the faster the reaction will take place.

2. Time: More time that current is allowed to pass, more products will be formed.

3. Charge on the ions: Look at the half-equations, and you can determine the mol ratios.

IB HL Objective

• 19.2.3 Describe the use of electrolysis in electroplating.


19.2.3 Describe the use of electrolysis in electroplating.

Electroplating: Cathode becomes coated in a layer of metal (negative electrode).

Cathode is where reduction takes place.

http://images.google.ae/imgres?imgurl=http://farm4.static.flickr.com/3152/3003863856_e9d5601266.jpg&imgrefurl=http://www.flickr.com/photos/chemheritage/3003863856/&usg=__MMS_hu1XVFwadH3HheLCXCIrwmk=&h=415&w=500&sz=79&hl=en&start=17&um=1&tbnid=PHrYg_1y6YfxDM:&tbnh=108&tbnw=130&prev=/images%3Fq%3Dsilver%2Belectroplating%26hl%3Den%26sa%3DG%26um%3D1

19.2.3 Describe the use of electrolysis in electroplating.

• Electroplating can be used to purify substances.

• For example, copper is used for electrical wiring, and it needs to be pure otherwise the resistance increases.

• So the positive electrode is impure copper and the negative electrode becomes pure copper.

http://images.google.ae/imgres?imgurl=http://www.beaducation.com/shop/images/copper_wire_spool.jpg&imgrefurl=http://www.beaducation.com/shop/index.php%3Fmain_page%3Dindex%26cPath%3D73_75&usg=__ocqzCEx7SDB5zyZkNK3l9_K1CI0=&h=500&w=500&sz=54&hl=en&start=1&um=1&tbnid=QVUd6qg8W_queM:&tbnh=130&tbnw=130&prev=/images%3Fq%3Dcopper%2Bwire%26hl%3Den%26sa%3DG%26um%3D1

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Topic 9 Oxidation and Reduction. IB Core Objective 9.1.1 Define oxidation and reduction in terms of electron loss and gain. Define: Give the precise meaning